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I needed to replace a solution in the equation, to see if it was correct.

The two solutions of this equation: $log_5{(x+3)} + log_5{(12x + 1)} = 3$,

are: {${\frac{-61}{12}, 2}$}

The solution $2$, satisfy the equation, but with $\frac{-61}{12}$ is the problem.

My development to replace in a image:

enter image description here

In step 2, I specifically marked it with a circle. My teacher says that it does not satisfy, since there are no negative logarithms, but if I continue to develop the equation, it remains as positive and if it satisfies, then who is wrong?

ESCM
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    $logA+logB=log(AB)$ is only true if $A>0$ and $B>0$ – imranfat Apr 26 '18 at 20:34
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    Please multiply the figure by $i$. – rafa11111 Apr 26 '18 at 20:35
  • And what is $A < 0$ and $B < 0$ ? so $ (- * - = + )$ – ESCM Apr 26 '18 at 20:35
  • What ? why $i$? – ESCM Apr 26 '18 at 20:36
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    In your course, the logarithm of a negative number does not exist. The logarithm has to exist in order to apply the standard rules of logarithms. Rafa's comment (please ignore!) is not going to change that – imranfat Apr 26 '18 at 20:36
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    In addition, you may want to resort to a graphing tool to graph $y=log_5{(x+3)} + log_5{(12x + 1)} $ and look at the graph to see where it crosses the line $y=3$. There will be only one solution – imranfat Apr 26 '18 at 20:39
  • "but if I continue to develop the equation"... you can't "continue to develop" if you are at point where something is impossible. A unicorn is a myth but cheese is not. If I take a unicorn, milk it and make cheese, I can't claim the end result, cheese, is okay. I can't have cheese, because I had to use a unicorn to get there. And unicorns don't exist. In the same way you can't add two logarithms of negative numbers together and get a logarithm of a positive number. You used negative numbers to get there, and logs of negatives don't exist. – fleablood Apr 26 '18 at 23:17
  • "Please multiply the figure by i." "What ? why i?". The guy is making a joke. He is saying "please rotate the image counterclockwise by 90 degress". If you multiply a complex number by $i$ the resulting (when graphed on the complex plane) is a 90 degree rotation of the original number. – fleablood Apr 26 '18 at 23:21

1 Answers1

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The conditions of the equation are

$$x+3>0$$ and $$12x+1>0.$$

which could be replaced by $$x>\frac {-1}{12} $$

but $$\frac {-61}{12}<\frac {-1}{12} $$ and is not a solution.

hamam_Abdallah
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